Thread: stacking a deck of cards

Here comes another one of my questions....
I'm placing it in the mathematics section because I want you all to address this from the standpoint of "in theory" and not "in practice". I'll explain more on that later.


So here is the problem;
Let's say you have as many decks of cards as you'll ever need. What you do is place one card on top of another so that the top card is just at the tipping (balance) point. Then you place those two cards on a third so that the top two are at the tipping point. Then place those three on a fourth... etc. Do this such that all the top cards are on the verge of tipping.

I'll attempt to "draw" a picture of this (I have a feeling it won't look very good so bare with me...):

__________xxxxxxxxxx
_____xxxxxxxxxx
__xxxxxxxxxx
xxxxxxxxxx

The question is;
Is it possible to stack the cards this way so that the top card is as far as you like to the right of the bottom card or is there a limit?


As far as the "in theory" vs. "in practice" - don't worry about things such as the curvature of Earth, inability to lift an infinite stack of cards, yada yada. Feel free to imagine a flat-infinite Earth etc. Nothing tricky here except possibly the math.

For the experimentalists in the forum and those that are not comfortable with the math, feel free to play with a deck of cards and rely on your intuition. For the mathematically-savvy, see if you can mathematically justify your answer.

Cheers,
william

The top card must have at least 50% over the stack, or base card, as must EVERY other card in the stack.


A guy goes to a builders office to pay a bill, when he looks at it he says "That's Nine times what it should be!" the builder merely crosses off the first digit which corrects the amount. An integer number of dollars is paid, what was the lowest the bill could have been before and after correction?.

Originally Posted by billco

The top card must have at least 50% over the stack, or base card, as must EVERY other card in the stack.


A guy goes to a builders office to pay a bill, when he looks at it he says "That's Nine times what it should be!" the builder merely crosses off the first digit which corrects the amount. An integer number of dollars is paid, what was the lowest the bill could have been before and after correction?.

45 and 5?

Correct, but not the answer I expected, I should have said the bill starts as 4 odd digits.

Sorry Will, I er inadvertantly snatched your thread....

Here, have it back!

Originally Posted by william

The question is;
Is it possible to stack the cards this way so that the top card is as far as you like to the right of the bottom card or is there a limit?

If I'm understanding you right, it sounds like eventually the center of mass of the stack will get too far away from the center of the base and it will fall.

Having just tried the cards out I am amazed at the answer I got, and that two sets of cards gave two different answers!

But how much error?

Hopefully william won't mind if I give this a little kick. I quite like this problem.

Originally Posted by Scifor Refugee

If I'm understanding you right, it sounds like eventually the center of mass of the stack will get too far away from the center of the base and it will fall.

They're all balanced.

The card 1 (the top card) is placed so it's centre of mass is on the edge of card 2, balancing it 'on the brink' of ruin, but still balanced. Cards 1 and 2 are then placed on card 3 so the are balanced perfectly on it's edge, and so on.

Note there is no sneezing allowed during this problem.

What you should work out is after you have n cards placed, how far to the right is card #1 from the centre of mass of this carefully stacked n cards? e.g. after 1 card, the rightmost edge of card #1 is 1/2 a length from the centre of mass (this example was hard work).

There is no limit - as the harmonic series diverges

Originally Posted by shmoe

They're all balanced.

The card 1 (the top card) is placed so it's centre of mass is on the edge of card 2, balancing it 'on the brink' of ruin, but still balanced. Cards 1 and 2 are then placed on card 3 so the are balanced perfectly on it's edge, and so on.

But eventually you'll get to a point where the stack extends 10 meters horizontally. The center of mass will be in the horizontal center of the stack, 5 meters out from the 4 cm x 7 cm (or whatever) base. Could that be stable? Or am I missing something?

Originally Posted by Scifor Refugee

But eventually you'll get to a point where the stack extends 10 meters horizontally. The center of mass will be in the horizontal center of the stack, 5 meters out from the 4 cm x 7 cm (or whatever) base. Could that be stable? Or am I missing something?

They aren't evenly spaced. Let's give them a length of 1.

Card #1 puts it's centre on the edge of card #2 and hangs out 1/2 over the edge.

The centre of mass of these two cards, measured from the left edge of card #2, is (1/2+1)/2=3/4. This centre of mass is placed on the edge of card #3 and so card #2 hangs 1/4 over the edge of card #3. Card #1 is still overhanging 1/2 over card #2, so is 1/2+1/4=3/4 over the edge of card #3.

and so on. Does that make sense? The overhang gets smaller and smaller to keep things balanced.

!/2+1/4+1/6+1/8+1/10+1/12+1/14+1/16.........

Which I think is an infinite series

Against all practical judgement, infinite overhang, infinite number of cards.

Originally Posted by river_rat

There is no limit - as the harmonic series diverges

Originally Posted by billco

!/2+1/4+1/6+1/8+1/10+1/12+1/14+1/16.........

Which I think is an infinite series

Against all practical judgement, infinite overhang, infinite number of cards.

Both correct!

Originally Posted by billco

Against all practical judgement, infinite overhang, infinite number of cards.

A more practical viewpoint for you, given any finite distance, you can surpass that with a finite number of cards.

A big hindrance to building something like this is the very slow rate of growth of the harmonic series. 1+1/2+1/3+...+1/n is about log(n), so if we wanted to go past say 10 card lengths, we'd need

(1/2)*(1+1/2+...+1/n)~10

or

log(n)~20

or n~exp(20) whcih is about 485,165,195 cards. That's starting to get rather high.


Note that more accurate approximations to the partial sums of the harmonic series can get you closer to the actual number of cards required to go out a given distance (without having to calculate these partial sums exactly).